Physics Thermodynamics
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New question posted
a month agoNew answer posted
a month agoContributor-Level 10
T = constant
P = constant
PV = nRT
PdV = nRdT
PdV + VdP = 0
ΔV = nRΔT/P
dV = (-)VdP/P
|ΔV| = V (ΔP/P)
V/P ΔP = nRΔT/P
ΔT = V/nR ΔP
C = V/nR T/P = 300/2 = 150
New answer posted
a month agoContributor-Level 10
ΔQ = heat supplied
ΔW = work done
ΔU = change in internal energy
(i) adiabatic (B) Δθ = 0
(ii) isothermal (D) ΔU = 0
(iii) isochoric (A) ΔW = 0
(iv) isobaric (C) ΔU ≠ 0, ΔW ≠ 0, ΔQ ≠ 0
New answer posted
a month agoContributor-Level 10
n? T? + n? T? = nT
⇒ (0.1) (200) + (0.05) (400) = (0.15)T
⇒ T = 266.67
New answer posted
a month agoContributor-Level 10
Mono atomic → Cv = 3R/2, Cp = 5R/2
Di-atomic → Cv = 5R/2, Cp = 7R/2
(Rigid)
Di-atomic → Cv = 7R/2, Cp = 9R/2
(Non-Rigid)
Tri-atomic → Cv = 3R, Cp = 4R
(Rigid)
New answer posted
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